Problem: Let $a$ and $b$ be positive real numbers.  Find the minimum value of
\[a^2 + b^2 + \frac{1}{(a + b)^2}.\]
Explanation: Let $s = a + b.$  By QM-AM,
\[\sqrt{\frac{a^2 + b^2}{2}} \ge \frac{a + b}{2} = \frac{s}{2}.\]Then $\frac{a^2 + b^2}{2} \ge \frac{s^2}{4},$ so $a^2 + b^2 \ge \frac{s^2}{2}.$  Hence,
\[a^2 + b^2 + \frac{1}{(a + b)^2} \ge \frac{s^2}{2} + \frac{1}{s^2}.\]By AM-GM,
\[\frac{s^2}{2} + \frac{1}{s^2} \ge 2 \sqrt{\frac{s^2}{2} \cdot \frac{1}{s^2}} = \sqrt{2}.\]Equality occurs when $a = b$ and $s^2 = \sqrt{2}.$  The numbers $a = b = 2^{-3/4}$ satisfy these conditions.

Therefore, the minimum value is $\boxed{\sqrt{2}}.$